Find the coordinates of the point \( R \) on the line segment joining the points \( \mathrm{P}(-1,3) \) and \( \mathrm{Q}(2,5) \) such that \( \mathrm{PR}=\frac{3}{5} \mathrm{PQ} \).
Given:
The line segment joining the points \( \mathrm{P}(-1,3) \) and \( \mathrm{Q}(2,5) \) such that \( \mathrm{PR}=\frac{3}{5} \mathrm{PQ} \).
To do:
We have to find the coordinates of the point \( R \).
Solution:
Let $R(x, y)$ be the point which divides the line segment joining the points $P(-1,3)$ and $Q(2,5)$
Given,
$P R=\frac{3}{5} P Q$
$\frac{P Q}{P R}=\frac{5}{3}$
$\frac{P R+R Q}{P R}=\frac{5}{3}$ (Since $PQ=PR+RQ$)
$1+\frac{R Q}{P R} =\frac{5}{3}$
$\frac{RQ}{P R}=\frac{5}{3}-1$
$=\frac{5-3}{3}$
$=\frac{2}{3}$
This implies,
$R Q: P R =2: 3$
$P R: R Q=3: 2$
Therefore,
$R(x, y)$ divides the line segment joining the points $P(-1,3)$ and $Q(2,5)$ in the ratio $3: 2$.
Using section formula, we get,
$(x,y)=(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}ny_{1}}{m+n})$
$(x, y)=(\frac{3(2)+2(-1)}{3+2}, \frac{3(5)+2(3)}{3+2})$
$=(\frac{6-2}{5}, \frac{15+6}{5})$
$=(\frac{4}{5}, \frac{21}{5})$
Therefore, the coordinates of the point \( R \) are $(\frac{4}{5}, \frac{21}{5})$.
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